How much energy is required to ionize a hydrogen atom containing an electron in the n 4n 4 level?

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What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom?

• $\pu{−3.49 * 10^{−17} J}$
• $\pu{−1.36 * 10^{−19} J}$
• $\pu{+2.18 * 10^{−18} J}$
• $\pu{+1.36 * 10^{−19} J}$

I assume that the way to do this is to start from the Rydberg formula,

$$ E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right), $$

set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to

$$ E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}. $$

I think you are asking the amount of energy input needed to bump the electron up from #n = 1# to #n = 3#... You may also mean the energy input needed to ionize hydrogen atom if the electron is in #n = 1, 2,# or #n = 3#...

Either way, the idea is to use the Rydberg equation (or some form of it), which describes the difference in energy for electronic transitions in the hydrogen atom:

#DeltaE = -R_H(1/n_f^2 - 1/n_i^2)#

It just so happens that #R_H# is defined to be the magnitude of the ground-state energy of #"H"# atom, #-"13.61 eV"#. #n_f# is the energy level of the destination state, and #n_i# is the energy level of the beginning state.

ELECTRONIC TRANSITION ENERGY

In this case, we have #n_i = 1# and #n_f = 3#, since we start on #n = 1# and end on #n = 3#. So...

#color(blue)(DeltaE) = -"13.61 eV"(1/3^2 - 1/1^2)#

#=# #color(blue)(ul"12.10 eV")#

In #"J"#, we would use the amount of #"J"# in #"1 eV"#:

#12.10 cancel("eV") xx (1.602 xx 10^(-19) "J")/(cancel"1 eV") = color(blue)(ul(1.938 xx 10^(-18) "J"))#

This should make physical sense. Electronic excitations (to higher #n#) require energy input with respect to the system.

Also, this number should be somewhat familiar; the electron has a charge of #-1.602 xx 10^(-19) "C"#, and #"1 C"cdot"V" = "1 J"#, so pushing an electron through a #"1 V"# potential difference involves #"1 eV"# of energy per electron.

IONIZATION ENERGIES

And if you somehow mean, what are #IE (n=1)#, #IE (n=2)#, and #IE(n=3)#, i.e. the ionization energies for #n = 1, n = 2, n = 3#, we simply use a single state from the Rydberg equation to get the ground-state energies:

#E_1 = -"13.61 eV"/1^2 = -"13.61 eV"# (at #n = 1#)

#E_2 = -"13.61 eV"/2^2 = -"3.40 eV"# (at #n = 2#)

#E_3 = -"13.61 eV"/3^2 = -"1.51 eV"# (at #n = 3#)

From Koopman's approximation theorem, we say that the ionization energy from an orbital containing one electron is approximately the same magnitude as the energy of that orbital, because it should take the electron past the #"0 eV"# threshold it needs to escape the atom.

So, the ionization energies would then be:

#"IE"(n = 1) = color(blue)(ul(+"13.61 eV"))#

#"IE"(n = 2) = color(blue)(ul(+"3.40 eV"))#

#"IE"(n = 3) = color(blue)(ul(+"1.51 eV"))#